1st Year Physics Lab Manual Experiment E3 (University of Colorado)

E3.1 Lab E3: The Wheatstone Bridge Introduction The Wheatstone bridge is a circuit used to compare an unknown resistance with a known resistance. The bridge is commonly used in control circuits. For instance, a temperature sensor in an oven often consists of a resistor with a resistance that increases with temperature. This temperature-dependent resistor is compared with a control resistor (outside the oven) to control a heater and maintain a set temperature. A schematic of a Wheatstone bridge is shown below:

Vo A

Ia

I

I

R1

R2

G

C Rk

D Rx

B

The unknown resistor is Rx, the resistor Rk is known, and the two resistors R1 and R2 have a known ratio R 2 R 1 , although their individual values may not be known. A galvanometer G measures the voltage difference VAB between points A and B. Either the known resistor Rk or the ratio R 2 R 1 is adjusted until the voltage difference VAB is zero and no current flows through G. When VAB = 0, the bridge is said to be “balanced�.

Ib

Fig.1. Schematic of a Wheatstone Bridge Since VAB = 0, the voltage drop from C to A must equal the voltage drop from C to B, VCA = VCB. Likewise, we must have VAD = VBD. So we can write,

(1)

I a R1 = I b R k

(2)

Ia R2 = Ib R x .

Dividing (2) by (1), we have (3)

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R2 Rx = R1 R k

,

Rx = Rk

R2 .

R1

E3.2 Thus, the unknown resistance Rx can be computed from the known resistance Rk and the known ratio R 2 R 1 . Notice that the computed Rx does not depend on the voltage Vo; hence, Vo does not have to be very stable or well-known. Another advantage of the Wheatstone bridge is that, because it uses a null measurement, (VAB = 0), the galvanometer does not have to be calibrated.

In practice, the Wheatstone bridge is seldom used merely to determine the value of a resistor in the manner just described. Instead, it is usually used to measure small changes in Rx due, for instance, to temperature changes or the motion of microscopic defects in the resistor. As an example, suppose Rx = 106 Ω and we wanted to measured a change in Rx of 1Ω, resulting from a small temperature change. There is no ohmmeter which can reliably measure a change in resistance of 1 part in a million. However, the bridge can be set up so that VAB = 0 when Rx is exactly 106 Ω. Then any change in Rx , ∆R x ,would result in a non-zero VAB, which, as we show below, is proportional to ∆R x .

You would not weigh a cat by weighing a boat with and without a cat on board.

Likewise, you would not want to measure very small changes in Rx by measuring Rx with and without the change. Instead, you want to arrange things so that the change in Rx, ∆R x , is the entire signal. The bridge serves to “balance out” the signal due Rx, leaving only the signal due to ∆R x .

To show that VAB ∝ ∆R x ,we consider Fig.1, and note that VCD = Vo . We assume that the galvanometer is a perfect voltmeter, so that no current flows through it, even when the bridge is not balanced. We also assume that the bridge has been balanced with the sample resistance at an initial value of Rxo , so that R xo = R k ( R 2 R 1 ) . Then we consider what happens to VAB when the sample resistor is changed by a small amount to a new value R x = R xo + ∆R x .

Applying Kirchhoff’s Voltage Law and Ohm’s Law to the upper and lower arms of the bridge, we have (4)

Vo = I a ( R 1 + R 2 ) = I b ( R k + R x ). .

We are trying to find VAB, which we can relate to Ia and Ib. (5)

VAB = VA − VB = ( VC − VB ) − ( VC − VA ) = I b R k − I a R 1 .

We can use (4) to solve for Ia and Ib and then substitute for Ia and Ib in (5), (6)

VAB

= Vo

R1 Rk . − Vo R1 + R 2 Rk + Rx

This equation shows how VAB depends on Rx. Notice that eqn.(6) yields VAB = 0 when R x = R k ( R 2 R 1 ) . To see how much VAB changes when Rx changes from Rxo to R xo + ∆R x , we write

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E3.3 ⎛ dV ⎞ Rk ≅ ⎜ AB ⎟ ⋅ ∆R x = − Vo ⋅ ∆R x .

2 ⎝ dR x ⎠ R = R ( R + R ) k xo x xo We wrote eqn.(7) by regarding VAB as a function of Rx, and remembering (from df Calculus) that if f = f ( x), then δf = ⋅ δx . dx Substituting R xo = R k ( R 2 R 1 ) into (7) yields (8) − Vo R 12 − Vo − Vo R k ⋅ ∆R x . ⋅ ∆R x = ⋅ ∆R x = ∆VAB = 2 2 2 R R + R R R ( ) 2⎞ ⎛⎜ R + R k 1 2 R k ⎛⎜ 1 + 2 ⎞⎟ k R ⎟⎠ R1 ⎠ ⎝ ⎝ k 1 ∆VAB

(7)

Finally, we remember that VAB,initial = 0, so the change in VAB is the same as VAB, and we have 2

(9)

VAB

⎛ R 1 ⎞ ∆R x = Vo ⎜ . ⎟ ⎝ R1 + R 2 ⎠ R k

We are done. We have shown that, when the bridge is balanced, any small change in Rx will produce a VAB proportional to that change.

Experiment We will use a slide-wire Wheatstone bridge, in which the two resistors R1 and R2 are two portions of a single, uniform Ni-Cr wire. Electrical contact is made at some point along the wire by a sliding contact (this contact corresponds to point A). The two portions of the wire on either side of the contact have resistances R1 and R2, and the ratio R 2 R 1 is the same as the ratio of the lengths of the two portions of wire, L 2 L1 . The lengths are readily measured with a meter stick which the wire rests upon.

A digital multimeter (DMM) in voltage mode will serve as the galvanometer.

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E3.4

10 Ω resistor

6 V power supply Rx (unknown)

Rk decade box B Ni-Cr wire on meter stick C

D A, sliding contact L1

L2

DMM Fig. 2. Physical Layout of the Wheatstone Bridge.

The 10Ω resistor in series with an adjustable power supply serves to limit the current through the bridge to less than 1A. (Higher currents might over-heat components of the bridge.) Set the voltage from the power supply to approximately 6 volts with the “Coarse” adjust voltage knob.

The known resistor Rk is adjustable and can be set to any value from 1Ω to 999Ω in 1Ω steps with a decade resistance box, which is accurate to 0.02%. The Ni-Cr wire has a total resistance of about 2Ω. The sliding contact is spring loaded; you have to push it down to make contact to the wire at point A. There are two buttons on the sliding contact; push one or the other, but not both, to contact the wire.

Wire to sliding contact. sliding contact nichrome wire

Adjust the position of the sliding contact to balance the bridge (zero reading on the DMM).

For Part 1 of the lab, the unknown resistor Rx is one of 5 coils of wire, numbered 1 through 5, mounted on a board. The lengths of the wires, their composition, and their Fall 2004

E3.5 gauge number are printed on the board. The gauge number is a measure of the thickness of the wire. 22 gauge wire has a diameter of 0.644 mm; 28 gauge wire has a diameter of 0.321 mm. For part 2 of the lab, Rx is a stand-alone coil of copper.

The resistance R of a wire is related to its length L, its cross-sectional area A, and its resistivity ρ by the formula

R=ρ

(10)

L .

A

The resistivity ρ of a material depends on composition, on defect structure, and on the temperature. For metals, ρ is approximately constant at very low temperatures (T < 100K) and increases approximately linearly with temperature (measured in K) at high temperatures.

At T = 20C, the resistivity of pure, defect-free, . × 10 −8 Ω ⋅ m . (RT copper is ρ RT = 1678 stands for room temperature.)

ρ

RT=293K

T ( oK)

Procedure Part 1: Resistivity of Copper

In this section, you will use the bridge to make precise measurements of the resistance of each of the 5 coils of wire on the coil board. First, however, use the digital multimeter (DMM) to make an approximate measurement of the resistance of each coil.

Also use the DMM to see how the resistance of the decade box changes when you turn the knobs. You must temporarily disconnect the coil board and the decade box from the bridge when testing them with the DMM. With the DMM, measure the resistance of each of the five coils to the nearest 0.1 Ω. The resistance of the wire leads used to connect the DMM to the coils is not negligible, so first use the DMM to measure the resistance of the two wire leads in series, and then subtract this lead resistance from your measurements. Check yourself: the smallest coil resistance is below 1 ohm!

Now familiarize yourself with the decade resistance box. Adjust the knobs for 15 ohms and verify with the DMM that the value is indeed 15 ohms plus the resistance of

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E3.6 the wire leads. For this measurement, disconnect the decade box from the bridge circuit and connect the DMM directly to the decade box.

Now use the bridge to measure the resistance of each of the 5 coils. Before connecting the power supply to the bridge, carefully check that all the connections are correct. Select a coil, attach it to the bridge, and set the decade box resistance Rk to be as near to Rx as possible (you know Rx roughly from your DMM measurements). Balance the bridge by moving the sliding contact along the wire while watching the DMM. Remember that when using the galvanometer you should always begin with button 1, then button 2, and finally button 3. With the bridge balanced, measure L1 and L2, and compute

Rx = Rk

(11)

R2 L = Rk 2 .

R1 L1

Repeat this procedure for the other 4 coils.

For each of these coils, compute the resistivity, using your measured resistances, the data printed on the coil board, and eq’n (10). Make a table with the headings: coil #, R from DMM, R from bridge, and computed resistivity. Four of the 5 sample coils are made of copper (Cu). For these four coils, compute the average resistivity, and the uncertainty of the average ( σ mean ). Compare your average value with the known value.

Coil #5 is made of a copper-nickel alloy. What is the ratio ρ Cu − Ni ρ Cu ?

Part 2. Temperature dependence of resistivity.

For this part, the unknown Rx is a copper wire coil (the one which is not attached to the coil board). We will use eq’n(9) to measure the change in resistance of the coil when its temperature is changed by plunging it into an ice bath. Note that eq’n(9) can be written as 2

(9’)

VAB

⎛ L1 ⎞ ∆R x = Vo ⎜ ⎟ ⎝ L1 + L 2 ⎠ R k

Prepare an ice bath by filling a beaker with ice from the freezer (ask your TA where) and adding some water. Measure room temperature and the ice bath temperature with the digital thermometer.

Measure the coil resistance Rx with the DMM and set Rk equal to Rx, as nearly as possible. With Rx and Rk connected in the bridge, measure Vo = VCD with the DMM. See Fig. 2 to locate VCD. Note that VCD is not the power supply voltage.

For this bridge measurement, we must replace the galvanometer in the bridge circuit with the DMM, set to DC volts with the most sensitive range(400mV). We need to do this because the galvanometer does not behave as an ideal voltmeter, as assumed in the derivation of eq’n (9).

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E3.7 With the sample coil at room temperature, balance the bridge and compute the coil’s resistance at room temperature. Now without changing the position of the slide wire contact, place the coil into the ice bath and watch as the voltage VAB on the DMM changes. Record VAB after it reaches equilibrium. Use eq’n(9) to compute ∆R x , the change in resistance of the coil.

As a measure of how sensitive the resistivity is to temperature changes, we can compute the fractional change in the resistivity divided by the change in temperature, (12)

( ∆ρ ρ) = 1 ∆ρ = ∆T

ρ ∆T

1 ∆R .

R ∆T

This quantity, multiplied by 100%, is the % change per degree. From your measurements, compute the % change in resistivity per degree for copper, and compare this value with the value you expect if ρ ∝ T( o K) . [Hint: If ρ ∝ T( o K) , then ρ = C ⋅ T , where C is some constant, and ∆ρ = C ⋅ ∆T .]

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E3.8 Prelab Questions 1. Two copper wires, labeled A and B, are at the same temperature, but the temperature is unknown. Wire B is three times as long and has 1/3 the diameter of wire A. Compute the ratio RB/RA. 2. A 28 gauge copper wire is 10 meters long. What is its resistance at room temperature?

3. Name two advantages of a Wheatstone bridge over an ordinary ohmmeter.

4. What is the formula for the total resistance of two resistors in parallel? Consider two resistors R1 = 2 Ω and R 2 = 200 Ω . What is the total resistance of these two resistors in

parallel. Give your answer to the nearest ohm.

5. Consider the circuit shown in Fig.2 (not Fig.1) and assume that the galvanometer acts like an ideal voltmeter (with infinite internal resistance). Recall that the total resistance of the Ni-Cr wire is 2Ω. If Rk and Rx are both very, very large compared to 2Ω, how much current flows through the Ni-Cr wire?

6. Suppose an unknown resistance Rx is measured with the bridge circuit shown in Fig. 2 and the result is Rx = 8.65Ω. The 6V power supply is then replaced with an 10V supply and Rx is re-measured. What is the new measured value of Rx? 7. A sample of copper wire is 200m long and has a diameter of 0.150 mm. Its resistance is determined to be 88.0Ω. What is the resistivity of the copper in this wire?

8. (Counts as 3 questions) A Wheatstone Bridge such as in Fig. 1 has a Vo = 10.0V .

With Rk = 5.00Ω and the unknown resistor Rx at room temperature, the bridge is balanced with L1 = 34.0cm and L2 = 66.0 cm. What is the value of Rx? While still attached to the bridge, the unknown Rx is placed in an oven, which raises its temperature by 33.0o C. The value of VAB is then found to be 0.127V. What is the resistance change of the unknown, ∆Rx? What is the % change per degree of the resistivity of this sample?

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